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Volgvan
3 years ago
11

What is the average rate of change from x = −2 to x = 0?

Mathematics
1 answer:
astraxan [27]3 years ago
7 0
The rate of change is +2 or 2 over 1 or 2/1
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Evaluate the following expression when x = 3 and y = 4:
OLEGan [10]

\dfrac{ {3}^{2} +  {4}^{3}  }{2 + 3}  \\  =  \dfrac{9 + 64}{5}  \\  =  \dfrac{73}{5}  \\  = 14.6

Answer is C.

Hope this helps. - M
3 0
3 years ago
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............................
lilavasa [31]

Answer:

............................

Step-by-step explanation:

8 0
3 years ago
[IMAGE ATTACHED]<br> I don’t understand this please help.
rusak2 [61]

Answer:

  • z+5+101=180°( being straight angle)
  • z=180°-106°
  • z=74°

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5 0
3 years ago
Use an equation to find the value of k so that the line that passes through the given points has the given slope. (
shutvik [7]

The formula of a slope:

m=\dfrac{y_2-y_1}{x_2-x_1}

We have the points (4, -4) and (k, -1) and the slope m = 34.

Substitute:

\dfrac{-1-(-4)}{k-4}=34\\\\\dfrac{3}{k-4}=\dfrac{34}{1}\qquad|\text{cross multiply}\\\\34(k-4)=3\qquad|\text{use distributive property}\\\\34k-136=3\qquad|+136\\\\34k=139\qquad|:34\\\\k=\dfrac{139}{34}

3 0
3 years ago
Max makes and sells posters. The function p(x)= -10x^2 +200x -250, graphed below, indicates how much profit he makes in a month
viktelen [127]
Here is our profit as a function of # of posters
p(x) =-10x² + 200x - 250
Here is our price per poster, as a function of the # of posters:
pr(x) = 20 - x
Since we want to find the optimum price and # of posters, let's plug our price function into our profit function, to find the optimum x, and then use that to find the optimum price:
p(x) = -10 (20-x)² + 200 (20 - x) - 250
p(x) = -10 (400 -40x + x²) + 4000 - 200x - 250
Take a look at our profit function. It is a normal trinomial square, with a negative sign on the squared term. This means the curve is a downward facing parabola, so our profit maximum will be the top of the curve.
By taking the derivative, we can find where p'(x) = 0 (where the slope of p(x) equals 0), to see where the top of profit function is.
p(x) = -4000 +400x -10x² + 4000 -200x -250
p'(x) = 400 - 20x -200
0 = 200 - 20x
20x = 200
x = 10                         
p'(x) = 0 at x=10. This is the peak of our profit function. To find the price per poster, plug x=10 into our price function:
price = 20 - x
price = 10
Now plug x=10 into our original profit function in order to find our maximum profit:
<span>p(x)= -10x^2 +200x -250
p(x) = -10 (10)</span>² +200 (10) - 250
<span>p(x) = -1000 + 2000 - 250
p(x) = 750

Correct answer is C)
</span>
7 0
3 years ago
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